# Proof $\pi$ is transcendental without symmetric function theory

This is a crosspost of my question from MSE from 3 weeks ago which was bountied but has received no response.

For an algebra assignment, I was asked to do a literature review and write up a proof of the transcendence of $$\pi$$. All sources I found presented Lindemann's proof (the "Transcendence of $$\pi$$" by Steve Mayer is an example), which uses symmetric function theory (particularly, the fundamental theorem of elementary symmetric functions) as a critical component of its argument, and as far as I am aware, generalizations like the Lindemann-Weierstrass theorem are similarly reliant on symmetric function theory in their proofs. Before the assignment, I did not know any symmetric function theory (perhaps this is a fault of my undergraduate education), and I found it the hardest part of the proof to learn and understand.

Now, learning the basics of symmetric function theory was worthwhile and will probably be useful for me in the future, but it begs the question: Is there a proof of the transcendence of $$\pi$$ without symmetric function theory, and if not, why?

As brought up by Paramanand Singh in the MSE thread, a modification of Lindemann's argument that proves the same claim in the argument where symmetric function theory was used originally (a certain polynomial constructed in the proof has integer coefficients) would suffice, though I'd also accept a "higher-tech" proof that is a different argument entirely, if you know of any that exist.

• As someone with a liking for symmetric function theory, you've managed to make me want to look (again) at the traditional proof! Commented Mar 2 at 23:10
• The proof of Gelfond-Schneider theorem, (as surveyed e.g. in appendix to Lang's Algebra), does not use symmetric functions. It's much less elementary than the usual expositions of Hermite-Lindemann's proofs, but on the other hand, it feels more natural and less ad hoc (to me). Commented Mar 3 at 10:02
• This is not an answer to your stated question, but here's a philosophical reason that symmetric functions show up. You have a polynomial equation with rational coefficients that is allegedly satisfied by π, and as always, the way you're going to get a contradiction is to find an integer strictly between 0 and 1. So you're going to need to argue in the world of rational numbers. Now, any symmetric function (with rational coefficients) of the roots of a polynomial with rational coefficients is going to be a rational number. Commented Mar 3 at 18:43
• (continued) So if you imagine that you were Lindemann, trying to discover the first proof of this fact, it would be natural to examine this large "free supply" of rational numbers when you're searching for a contradiction. That is to say, it's not so much that symmetric functions are "necessary" for the argument; it's more that they're a natural arena to look for candidates for a contradiction. From this point of view, refusing to look at this arena strikes me as more like tying one hand behind your back than simplifying the argument. Commented Mar 3 at 18:43
• I accepted this $p$-adic proof of Lindemann-Weierstrass here mattbaker.blog/2015/03/20/… as the answer on my MSE question since it answered the spirit of what I was looking for, as for the letter of my question that can probably be answered by the experts more familiar with the field: does this $p$-adic argument bypass the symmetric function theory (or at least use it non-essentially), or is it used in important ways in the relied upon results? Commented Apr 30 at 16:39